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# A particle moves along a straight line such that its displacement at any time t is given by s = t3 - 6t2 + 3t + 4 metres The velocity when the acceleration is zero is (a)    3 ms-1        (b)    -12 ms-1         (c)    42 ms-2      (d)    -9 ms-1

7 years ago

(d)
Velocity, v=  ds/dt=3t^2  -12 t+3
Acceleration a = ds/dt = 6t - 12; For a = 0 we have, 0 = 6 t - 12 or t= 2s. Hence, at t = 2 s the velocity will be
v = 3 × 22 – 12 × 2 + 3 = -9 ms-1

3 years ago
(d) -9m/s we know that differentiation of displacement gives velocity i.e. 3t^2-12t+3.now we know that differentiation of velocity gives acceleration i.e.6t-12but here acceleration is zero;therefore,6t-12=0;t=2secnow put the value of `t` in velocity we get -9m/sThank You
2 years ago
2 years ago
ds/dt=v=3t^2-12t+3
d^2s/dt^2=a=6t-12
since a=0
therefore, 6t-12=0
6t=12
t=2
put t=2 in v
v= 3t^2-12t
v=-12m/s
=3(2^2)-12×2
V= -12m/s
2 years ago
Vineet Vilas kol …. BRO YOU HAVE BEEN PUT YOUR LEG IN DAL BRO YOU LEFT THE NUMBER 3 ON YOUR CALCULATION... SO CHECK IT ONCE.... AND THE CORRECT THE ANSWER TO THE OPTION D (-9)

11 months ago
hello student

since ,Velocity, v= ds/dt=3t^2 -12 t+3
Acceleration a = ds/dt = 6t - 12; For a = 0 we have, 0 = 6 t - 12 or t= 2s. Hence, at t = 2 s the velocity will be
v = 3 × 2^2 – 12 × 2 + 3 = -9 ms-1

Hope it helps
Thankyou Yash Chourasiya
10 months ago
Dear Student

Given
s= t3 − 6t2 + 3t + 4
v = ds/dt​
So v = 3t2 −12t + 3.............(1)
again differentiate, a = dv/dt​
6t − 12=0
t = 2s
Putting value of t in eq. (1)
v = 3(2)2 −12×2 +3
v = −9ms−1